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2 years ago

Which is the next logical step in balancing the given equation? 3CO(g) + Fe2O3(s) Fe(s) + 3CO2(g)

Chemistry

2 answers:

nikitadnepr [17]2 years ago

8 0

Answer : The correct option is, (C) Place the coefficient 2 in front of elemental iron.

Explanation :

Balanced chemical reaction : It is defined as the number of atoms of individual elements present on reactant side must be equal to the product side.

The given unbalanced chemical reaction is,

$3CO(g)+Fe_2O_3(s)\rightarrow Fe(s)+3CO_2(g)$

This chemical reaction is an unbalanced reaction because in this reaction, the number of atoms of iron are not balanced while all the atoms are balanced.

In order to balanced the chemical reaction, the coefficient 2 is put before the 'Fe'.

Thus, the balanced chemical reaction will be,

$3CO(g)+Fe_2O_3(s)\rightarrow 2Fe(s)+3CO_2(g)$

Elena-2011 [213]2 years ago

5 0

I believe that the answer is A.

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51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b

Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

$a = \sqrt{8} r$

$a = \sqrt{8} \times 144 pm$

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}$

density d = 19.41 g/cm³

Similarly; For a body-centered cubic structure

$r = \dfrac{\sqrt{3}}{4}a$

where;

r = 144

$144 = \dfrac{\sqrt{3}}{4}a$

$a = \dfrac{144 \times 4}{\sqrt{3}}$

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V 3.68 × 10⁻²³ cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}$

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

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